fluid_mechanics

6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids 317
E XAMPLE 9.10
Laminar Flow in an Annulus
GIVEN A viscous liquid
0.0045 N # 1r 1.18 10 3 kgm 3 ; m
sm 2 2 flows at a rate of 12 mls through a horizontal,
4-mm-diameter tube.
FIND 1a2 Determine the pressure drop along a l-m length of the
tube which is far from the tube entrance so that the only component
of velocity is parallel to the tube axis. 1b2 If a 2-mm-diameter rod is
placed in the 4-mm-diameter tube to form a symmetric annulus,
what is the pressure drop along a l-m length if the flowrate remains
the same as in part 1a2?
SOLUTION
(a) We first calculate the Reynolds number, Re, to determine
whether or not the flow is laminar. With the diameter
D 4 mm 0.004 m, the mean velocity is
and, therefore,
Re rVD
m
1000
V
Q
1p42D 112 ml s2110 6 m 3 ml2
2 1p4210.004 m2 2
0.955 ms
Since the Reynolds number is well below the critical value of
2100 we can safely assume that the flow is laminar. Thus, we can
apply Eq. 6.151, which gives for the pressure drop
¢p 8m/Q
pR 4
(Ans)
(b) For flow in the annulus with an outer radius r o 0.002 m
and an inner radius r i 0.001 m, the mean velocity is
and the Reynolds number [based on the hydraulic diameter,
D h 21r o r i 2 210.002 m 0.001 m2 0.002 m] is
Re rD hV
m
11.18 103 kgm 3 210.955 ms210.004 m2
0.0045 N # sm 2
810.0045 N # sm 2 211 m2112 10 6 m 3 s2
p10.002 m2 4
8.59 kPa
Q
V
p1r o 2 r i 2 2 12 10 6 m 3 s
1p2310.002 m2 2 10.001 m2 2 4
1.27 ms
11.18 103 kgm 3 2 10.002 m2 11.27 ms2
0.0045 N # sm 2
666
This value is also well below 2100 so the flow in the annulus
should also be laminar. From Eq. 6.156,
¢p 8m/Q
p
c r4 o r 4 i 1r 2 2
o r
ln1r o r i 2
i 2 2
1
d
so that
¢p 810.0045 N # sm 2 211 m2112 10 6 m 3 s2
p
e10.002 m2 4 10.001 m2 4
310.002 m22 10.001 m2 2 4 2 1
f
ln10.002 m0.001 m2
68.2 kPa
(Ans)
COMMENTS The pressure drop in the annulus is much larger
than that of the tube. This is not a surprising result, since to maintain
the same flow in the annulus as that in the open tube, the average
velocity must be larger (the cross-sectional area is smaller) and
the pressure difference along the annulus must overcome the shearing
stresses that develop along both an inner and an outer wall.
By repeating the calculations for various radius ratios, r ir o ,
the results shown in Fig. E6.10 are obtained. It is seen that the
pressure drop ratio, ¢p annulus¢p tube (i.e., the pressure drop in the
annulus compared to that in a tube with a radius equal to the outer
radius of the annulus, r o ), is a strong function of the radius ratio.
Even an annulus with a very small inner radius will have a pressure
drop significantly larger than that of a tube. For example, if the inner
radius is only 1100 of the outer radius, ¢p annulus¢p tube 1.28.
As shown in the figure, for larger inner radii, the pressure drop ratio
is much larger [i.e., p annulus p tube 7.94 for r ir o 0.50 as in
part (b) of this example].
________
Δp annulus
Δp tube
8
7
6
5
4
3
2
1.3
1.2
1.1
1
0 0.005 0.01
F I G U R E E6.10
(0.01, 1.28)
(0.50, 7.94)
1 0 0.1 0.2 0.3 0.4 0.5
r i /r 0